Define $f(x, y) = e^x + \dfrac{1}{y}$. Let $\vec{a} = (-2, 4)$ and $\vec{v} = \left( 0, 1 \right)$. Calculate $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h}$.
Answer: When a directional derivative has a direction that equals $(1, 0)$ or $(0, 1)$, it becomes a regular partial derivative. Because $v = (0, 1)$, the limit we want to find is the definition of $\dfrac{\partial f}{\partial y}$ evaluated at $(-2, 4)$. Therefore: $ \lim_{h \to 0} \dfrac{f \left( -2, 4 + h \right) - f(-2, 4)}{h} = \dfrac{\partial f}{\partial y}(-2, 4)$ $\begin{aligned} &\dfrac{\partial f}{\partial y} = \dfrac{-1}{y^2} \\ \\ &\dfrac{\partial f}{\partial y}(-2, 4) = \dfrac{-1}{16} \end{aligned}$ In conclusion, $ \lim_{h \to 0} \dfrac{f(\vec{a} + h \vec{v}) - f(\vec{a})}{h} = \dfrac{-1}{16}$.